设AG=a,BG=b,AE=x,ED=y
则a+b=x+y ①
2ax=by ②
由①得:a-x =y-b
两边平方得a2-2ax+x2=b2-2by+y2
把②代入得a2-2ax+x2=y2-4ax+b2
即(a+x)2=b2+y2,a+x= (√b)2+y2.
又∵b2+y2=CH2+CF2=FH2
∴a+x=FH
即DH+BF=FH
延长CB到M,使BM=DH,连结AM
∵Rt△ABM≌Rt△ADH
∴AM=AH,∠MAB=∠HAD
则∠MAH=∠MAB+∠BAH=∠BAH+∠HAD=90°
可证△AMF≌△AHF
则∠MAF=∠HAF
即∠HAF=1/2∠MAF=45°